We use the integral ∫(ydx) between limits √2 and 4.
So we replace ∫ydx by (1/3)∫((x^2-2)^(3/2)dx).
Let's try substitution:
let x=secø√2, dx=(√2)secøtanødø, x^2-2=2(secø)^2-2=2((secø)^2-1)=2(tanø)^2.
The integral becomes: (1/3)∫((√2(tanø))^3*√2secøtanødø). (1/3)(2√2)(√2)=4/3.
The limits are given by √2=secø√2, secø=1=1/cosø, so ø=0; 4=secø√2, ø=cos^-1(√2/4).
(4/3)∫((tanø)^4secødø)=(4/3)S where S=∫((tanø)^4secødø).
But (tanø)^4=((secø)^2-1)^2=((secø)^4-2(secø)^2+1) for all ø as a trig identity.
S=∫((secø)^5dø)-2∫((secø)^3dø)+∫(secødø).
Derivatives of secø and tanø are respectively secøtanø and (secø)^2.
So if p=secø+tanø then dp/dø=secø(tanø+secø)=psecø. Therefore dp/p=secødø.
Integrating we get ln(p)=ln(secø+tanø)=∫secødø.
Now let p=secøtanø, then dp/dø=(secø)^3+secø(tanø)^2=2(secø)^3-secø.
So dp=(2(secø)^3-secø)dø;
integrating: p=secøtanø=2∫(secø)^3dø-∫secødø=2∫(secø)^3dø-ln(secø+tanø), from which we get 2∫(secø)^3dø=secøtanø+ln(secø+tanø).
We now have to find ∫((secø)^5dø).
∫((secø)^5dø)=∫((secø)^3.(secø)^2dø).
Let u=(secø)^3, so du=3(secø)^2secøtanødø=3(secø)^3tanødø.
Let dv=(secø)^2dø, so v=tanø.
"d(uv)=vdu+udv" so ∫udv=uv-∫vdu. Applying this rule:
∫((secø)^5dø)=tanø(secø)^3-∫(3(secø)^3(tanø)^2dø)=tanø(secø)^3-∫(3(secø)^3((secø)^2-1)dø)=
tanø(secø)^3-3∫((secø)^5+3∫(secø)^3dø). We have the required integral on both sides of the equation: so, dividing through by 4 and substituting for ∫((secø)^3dø), we have:
∫((secø)^5dø)=(2tanø(secø)^3+3(secøtanø+ln(secø+tanø)))/8.
S=(2tanø(secø)^3+3(secøtanø+ln(secø+tanø)))/8-secøtanø-ln(secø+tanø)+ln(secø+tanø),
which simplifies to S=(2tanø(secø)^3-5secøtanø+3ln(secø+tanø))/8.
We need 4/3S=(2tanø(secø)^3-5secøtanø+3ln(secø+tanø))/6
Now we need to apply the limits: first ø=0:
S(0)=0 and ø=cos^-1(√2/4) so secø=4/√2=2√2 and tanø=√7.
(4/3)S(cos^-1(√2/4))=(2√7*16√2-10√14+3ln(2√2+√7))/6.
Definite integral becomes (22√14+3*1.7)/6=87.4166/6=14.5694 approx., which appears to be the area under the given curve between the given limits. I think!